H3C6H5O7(aq) + 3Na3HCO3(q)---Na3C6H5O7(aq) + 3H2O(l) + #CO2(g)
NaHCO3 = 1.00g
H3C6H5O5 = 0.76g
CO2= 0.5g
1) show that equivalence amount of cirtic acide for 1.00g of sodium bicarbonate is 0.76g
2)Calculate the theoretical yield of carbon dioxide from 1.00g of sodium bicarbonate
3) what is the actual yield
4) calculate the percentage
2 answers
what joins two lipids
I have corrected the typo in the equation below.
H3C6H5O7(aq) + 3NaHCO3(q)---Na3C6H5O7(aq) + 3H2O(l) + 3CO2(g)
1.
mols NaHCO3 = grams/molar mass = 1/84 = 0.0119
Convert mols NaHCO3 to mols H3C6H5O7 using the coefficients in the balanced equation. That's
0.0119 x (1 mol citric acid/3 mols NaHCO3) = 0.0110 x 1/3 = 0.003968
Now convert mols citric acid to grams. g = mols x molar mass = 0.003968 x 192.1 = 0.76
2.
Convert mols NaHCO3 to mols CO2. That is 0.0119 x (3 mols CO2/3 mols NaHCO3) = 0.0119
Then grams CO2 = mols CO2 x molar mass CO2 = 0.0119 x 44 = 0.524 g is theoretical yield (TY).
3.
The actual yield(AY) in the problem is given as 0.5 g
4.
% yield = (AY/TY)*100 = ?
H3C6H5O7(aq) + 3NaHCO3(q)---Na3C6H5O7(aq) + 3H2O(l) + 3CO2(g)
1.
mols NaHCO3 = grams/molar mass = 1/84 = 0.0119
Convert mols NaHCO3 to mols H3C6H5O7 using the coefficients in the balanced equation. That's
0.0119 x (1 mol citric acid/3 mols NaHCO3) = 0.0110 x 1/3 = 0.003968
Now convert mols citric acid to grams. g = mols x molar mass = 0.003968 x 192.1 = 0.76
2.
Convert mols NaHCO3 to mols CO2. That is 0.0119 x (3 mols CO2/3 mols NaHCO3) = 0.0119
Then grams CO2 = mols CO2 x molar mass CO2 = 0.0119 x 44 = 0.524 g is theoretical yield (TY).
3.
The actual yield(AY) in the problem is given as 0.5 g
4.
% yield = (AY/TY)*100 = ?
2 answers